 Solution

Introduce a constant Lagrange multiplier $\lambda$ for the fixed arc length. Up to an irrelevant constant term, the augmented functional isIt has no explicit $x$ dependence, so the Beltrami identity givesFor the hanging minimum choose $k>0$. Writing $y_0=-\lambda$, this becomes $1+y^{\prime 2}=(y-y_0{)}^2/k^2$. Its nonconstant smooth solution isOne may verify it without dividing by $y^{\prime }$ at the lowest point: the full Euler-Lagrange equation gives $(y+\lambda )y^{\prime \prime }=1+y^{\prime 2}$, which the expression satisfies everywhere. The branch with negative $k$ is the inverted stationary chain rather than the hanging minimum. The catenary minimizes the integral of height along a fixed-length curve, so the positive branch is also the global minimum, not merely a stationary candidate.